Dynamic equilibrium (I thought I understood now went blank again)

[scilearner]

Hello everyone,

 

It is funny that I thought I understood dynamic equilibrium really well but now I have gone blank again. I can do all the calculations and I understand

Le chatelier principle well. Here are the problems I have.

 

2A+B <----> C

 

* If forward reaction equals backward reaction wouldn't those two balance out and be zero. Is the reason why this doesn't happen is because both reactions are continuosly occuring. But please elaborate if you can.

 

*I checked so many sites on this topic and none of them really talks much about what happens to the moles in dynamic equilibrium. Can anyone tell me what happens to the moles in dynamic equilibrium. I understand that if the reaction occurs according to stoichiometric ratios there would be one way reaction. So roughly when we react 2 moles of A and 1 mole of B do we get like 0.8 moles of C.I'm just asking does something like this happen.

 

*If in equilibrium the reactants and products don't act according to their stoichiometric ratios why do we assume so in calculations. Do they always provide us with ICL data in equilibrium questions.

 

If anyone can take their time and at least answer one of these questions I would be very greatful. Thank you

[CaptainPanic]

Yes, if there is equilibrium, both reactions are occurring at the same time, and the net-effect is that there is no change. I don't know how to be more elaborate actually... so I'll go for an example

 

If in 1 second time, you have:

2 moles of A + 1 mole of B --> 1 mole of C

but also (in the same second):

1 mole of C --> 2 mole of A + 1 mole of B

 

then, one second later, a lot of reactions have happened, but nothing much changed.

 

Good examples of equilibrium are: water!

 

The following reactions happen all the time in water! In the sea, in your body, in beer, everywhere!

[ce]2 H2O -> OH^- + H3O^+[/ce]

and:

[ce]OH^- + H3O^+ ->2 H2O [/ce]

 

The average concentration in normal water (with a pH of 7) is:

10^-7 mol/liter of both OH- and H3O+

[hermanntrude]

the easiest way to visualise this is to have two tanks of water (imaginary or real), and two people, each with a bucket. They both scoop one bucket of water per second from one tank into the other. One person represents the forward reaction and the other represents the reverse reaction. The average water levels in the two tanks remain constant, however, water is moving in both directions.

 

Interestingly, if you add a catalyst to a dynamic equilibrium, it's just like asking the guys to scoop water faster. They BOTH scoop faster (the forward AND reverse reactions are sped up) and so the equilibrium is still maintained and the levels of water don't change.

 

Note also that if one tank has more water to begin with, it will stay that way. You can think of this as a bit like the equilibrium constant, which gives an idea of how far between reactants and products the equilibrium lies. It neednt be 50/50, in fact it rarely is.

 

all reactions are equilibria, but in some cases the equilibrium lies so far toward the products that we might as well pretend the reaction goes to completion, and sometimes the equilibrium lies so far to the reactants, we might as well pretend the reaction doesn't happen at all.

[CaptainPanic]

hermanntrude, do you have an example of a catalyst that speeds up reactions in an equilibrium, without changing the equilibrium constant (without moving the equilibrium)?

[scilearner]

Thank you Captain Panic and Hermantrude . I think I get it now. The 1 second example and the water bucket example were very helpful.. So with the water bucket example is this what happens

 

Intially one bucket was filled with water (reactants). First the water was poured it to another bucket very fast. Then as the water in the othe bucket increased the reverse reaction increases. Then it comes a stage where both people scoop only one cup of water at a second and this is equilibrium. Thanks again for both of you:-). I think I get it now.

[hermanntrude]

hermanntrude, do you have an example of a catalyst that speeds up reactions in an equilibrium, without changing the equilibrium constant (without moving the equilibrium)?

 

In an exam, the usual question is to give an equilibrium and ask what effect various changes would have on the concentration of one of the products or reactants. An example would be:

 

[ce]2H2O <=> H3O+ + OH-[/ce]

 

The question many students get wrong is "what would be the effect of adding a catalyst on the concentration of [ce]H3O+[/ce]?" Now I don't actually know of any catalysts for this reaction specifically, and it's kind of hard to think of an actual catalyst for an actual equilibrium, since people don't often use them, because they don't do much good, except in cases where the reaction rates are slow.

 

However, the accepted answer for the above question is "no change"

 

I suspect that this is actually a simplification and isn't entirely true, however, it's what is taught at degree level.

[CaptainPanic]

However, the accepted answer for the above question is "no change"

hermanntrude, I am afraid I do not understand what you wrote here (in your previous post)...

 

quoting from an older post of you:

 

Interestingly, if you add a catalyst to a dynamic equilibrium, it's just like asking the guys to scoop water faster. They BOTH scoop faster (the forward AND reverse reactions are sped up) and so the equilibrium is still maintained and the levels of water don't change.

 

I was wondering if you can show an example of a catalyst that'll make the scooping go faster (by whatever means, let's not get stuck in the metaphore)... "no change" does not seem a likely answer at this moment to me... the only thing I can come up with is an increase in temperature, which is not a catalyst.

[John Cuthber]

Catalysts do not, and can not, change the position of an equilibrium.

Show me a counter example and I will explain how to make it into a perpetual motion machine.

[hermanntrude]

hermanntrude, I am afraid I do not understand what you wrote here (in your previous post)...

 

quoting from an older post of you:

 

 

 

I was wondering if you can show an example of a catalyst that'll make the scooping go faster (by whatever means, let's not get stuck in the metaphore)... "no change" does not seem a likely answer at this moment to me... the only thing I can come up with is an increase in temperature, which is not a catalyst.

 

The reason that I can't easily provide you with an example is that examples of catalysts in equilibria are not usually quoted in the literature because there is no benefit in adding one, since it doesn't change the position of the equilibrium, only the speed at which it is reached.

[John Cuthber]

"hermanntrude, do you have an example of a catalyst that speeds up reactions in an equilibrium, without changing the equilibrium constant (without moving the equilibrium)?

"

That's all a catalyst can ever do so any actalyst will fit the bill.

For example the vanadium pentooxide used in the contact process for sulphuric acid.

[CaptainPanic]

Hmmm....

That all sounds reasonable. Anyways... I am afraid that I still have questions...

 

The equilibrium constant is (I copy pasted this from wikipedia):

 

Reaction:

[math] \alpha A +\beta B ... \rightleftharpoons \sigma S+\tau T ... [/math]

 

Equilibrium constant:

[math] K=\frac{{\{S\}} ^\sigma {\{T\}}^\tau ... } {{\{A\}}^\alpha {\{B\}}^\beta ...} [/math]

 

In that formula, for example {A} is the activities of species "A". (Activity is approximately the same as "concentration", and is exactly the same (by definition) in ideal cases). The activity is unaffected by the catalyst, and the catalyst itself is not present in the formula for the equilibrium.

 

Whenever I encounter the word "activity (coefficient)" in thermodynamics, my standard response is to stop doing thermodynamics calculations, and to open some handbook or google, and go hunting for data. I've never been able to determine an activity coefficient, and I have never really found the whole thing very practical for engineering.

 

Anyway, can we conclude that because catalysts cannot move an equilibrium, they cannot have any influence on the activity or on an activity coefficient? (Because this is troubling me a little... the activity coefficient shows how non-ideal a system is, and I can imagine that adding a catalyst can make something more or less ideal, thereby influencing the activity of other species?)

 

Hope I am still making sense... my goal is to learn some about this topic. It has never been my strong point. Thanks for any feedback.

[John Cuthber]

The equilibrium constant for a reaction can be calculated from the free energy change. The presense or absense of a catalyst can't change that.

The catalyst affects the rate of a reaction but, because of the conservation of energy, it is constrained to affect the reaction just as much in the oposite direction. If it affects the activity of one component then it must affect the activity of some othe component in order to restore the balance (actually there are exceptions to this but they are not strictly true catlysts).

[CaptainPanic]

Ah, cool... never realized that there was in a way an "activity balance", if you could call it like that. Thanks for the info / explanation. The world makes sense again.